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Real Transformer 1- There are flux leakages 2- Some power loss (copper, core losses) 3- Limited permeability Let’s see what effects these realities would.

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Presentation on theme: "Real Transformer 1- There are flux leakages 2- Some power loss (copper, core losses) 3- Limited permeability Let’s see what effects these realities would."— Presentation transcript:

1 Real Transformer 1- There are flux leakages 2- Some power loss (copper, core losses) 3- Limited permeability Let’s see what effects these realities would have on voltage and current relation of the primary and the secondary Q:Can we modify the ideal transformer model to take these realities into account? A: Yes

2 1-Flux leakage

3 With no flux leakage Now with flux leakage

4 Q: Is there a way to model the flux leakage: A: Yes, we can count for flux leakage using this equivalent circuit

5 Manipulations of formulas The same for secondary

6 2- Some power loss (copper, core losses) 2-1- Copper loss

7 2-2- Core (eddy and hysteresis) loss 3- Limited permeability In ideal transformer: That means in ideal transformer, if the current in the secondary is zero (no load), the current in the primary should be zero as well

8 In real transformer: That means in real transformer, if the current in the secondary is zero (no load), the current in the primary is not zero Magnetization current

9 Using phasor Ideal Transformer Model Real Transformer Model

10 Referred to the primary The original equivalent circuit Other Equivalent Circuits

11 Referred to the secondary The original equivalent circuit

12 Some Simplifications If we assume that the current in R c and X M would not change much if we change its place,

13 If we ignore the core losses and magnetization current These models are too simplified, the original model is too complicated, we usually work with these models Referred to the secondaryReferred to the primary

14 Determining the Parameters Open-circuit test Short-circuit test

15 Q: Given six values, how can we determine the parameters of the equivalent circuit From open-circuit test,

16

17 Short circuit test

18 Note: We cannot split the series impedance into primary and secondary components

19 In Summary Given: If the test is done on the primary For the equivalent circuit referred to the primary

20 Two Obtainable Equivalent Circuit To obtain the equivalent circuit referred to secondary, Divide the parameters by the turn ratio square

21 Example

22 Equivalent circuit Referred to the primary Equivalent circuit Referred to the secondary Given: 8000/240


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